What will be the equation of that chord of ellipse $\frac{{{x^2}}}{{36}} + \frac{{{y^2}}}{9} = 1$ which passes from the point $(2,1)$ and bisected on the point

Let the tangent and normal at the point $(3 \sqrt{3}, 1)$ on the ellipse $\frac{x^2}{36}+\frac{y^2}{4}=1$ meet the $y$-axis at the points $A$ and $B$ respectively. Let the circle $C$ be drawn taking $A B$ as a diameter and the line $x =2 \sqrt{5}$ intersect $C$ at the points $P$ and $Q$. If the tangents at the points $P$ and $Q$ on the circle intersect at the point $(\alpha, \beta)$, then $\alpha^2-\beta^2$ is equal to

An ellipse is described by using an endless string which is passed over two pins. If the axes are $6\ cm$ and $4\ cm$, the necessary length of the string and the distance between the pins respectively in $cm$, are

Let $F_1$ & $F_2$ be the foci of an ellipse $\frac{{{x^2}}}{4} + \frac{{{y^2}}}{9} = 1$ such that a ray from $F_1$ strikes the elliptical mirror at the point $P$ and get reflected. Then equation of angle bisector of the angle between incident ray and reflected ray can be 

The radius of the circle having its centre at $(0, 3)$ and passing through the foci of the ellipse $\frac{{{x^2}}}{{16}} + \frac{{{y^2}}}{9} = 1$, is